FFHow/OlympiadBench

1id	question	solution	final_answer	context	image	modality	difficulty	is_multiple_answer	unit	answer_type	error	question_type	subfield	subject	language20	"(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form3 4$$5\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))6\tag{2}7$$8 9For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,10 11$$12\begin{aligned}13& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\14& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y15\end{aligned}16\tag{3}17$$18 19where dots represent time derivatives.20 21Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:22 23$$240=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta25\tag{4}26$$"	['We can write down the equations of motion, multiplying the second one by $i$ :\n\n$$\n\\begin{aligned}\n& 0=\\ddot{x}+2 \\Omega \\dot{y}-\\Omega^{2} x \\\\\n& 0=\\ddot{y}-2 \\Omega i \\dot{x}-\\Omega^{2} i y\n\\end{aligned}\n\\tag{19}\n$$\n\nWe can add these equations together to obtain\n\n$$\n0=(\\ddot{x}+i \\ddot{y})+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2}(x+i y)=\\ddot{\\eta}+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2} \\eta\n\\tag{20}\n$$\n\nWe note that\n\n$$\n\\dot{y}-i \\dot{x}=-i(\\dot{x}+i \\dot{y})=-i \\eta\n\\tag{21}\n$$\n\nThen\n\n$$\n0=\\ddot{\\eta}-2 i \\Omega \\dot{\\eta}-\\Omega^{2} \\eta\n\\tag{22}\n$$']		"4. A complex dance 27 28In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.29 30Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:31 32$$33e^{i \theta}=\cos \theta+i \sin \theta34\tag{1}35$$"	[]	Text-only	Competition	False				Theorem proof	Mechanics	Physics	English361	(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.	['We can multiply both sides of the answer to part (e) by $\\hbar$ and use $E=\\hbar \\omega$ and $p=\\hbar k$, we have\n\n$$\nE=\\frac{p^{2}}{2 m}\n\\tag{33}\n$$\n\nA classical momentum has $p=m v$, so this gives the classical energy for a free particle (i.e., one without a potential):\n\n$$\nE=\\frac{1}{2} m v^{2}\n\\tag{34}\n$$']		"4. A complex dance 37 38In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.39 40Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:41 42$$43e^{i \theta}=\cos \theta+i \sin \theta44\tag{1}45$$46Context question:47(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form48 49$$50\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))51\tag{2}52$$53 54For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,55 56$$57\begin{aligned}58& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\59& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y60\end{aligned}61\tag{3}62$$63 64where dots represent time derivatives.65 66Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:67 68$$690=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta70\tag{4}71$$72Context answer:73\boxed{证明题}74 75 76Context question:77(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.78 79Plugging in this guess, what must $\lambda$ be?80Context answer:81\boxed{$\lambda=i \Omega$}82 83 84Context question:85(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.86 87This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.88Context answer:89\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}90 91 92Context question:93(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by94 95$$96\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}97\tag{5}98$$99 100A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation.""101Context answer:102\boxed{$\omega=-i k^{2} a$}103 104 105Context question:106(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by107 108$$109i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}110\tag{6}111$$112 113Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?114Context answer:115\boxed{$\omega=\frac{\hbar k^{2}}{2 m}$}116"	[]	Text-only	Competition	False				Theorem proof	Mechanics	Physics	English1172	"(b) Laplace's equation is a second order differential equation118 119$$120\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0121\tag{8}122$$123 124Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.125 126Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.127 128Hint: Consider $\phi=\phi_{1}-\phi_{2}$."	"[""Suppose that we have two potentials satisfying the Laplace's equation and the boundary conditions. $\\nabla^{2} \\phi_{1}=0, \\nabla^{2} \\phi_{2}=0$ and $\\phi_{1}=\\phi_{2}$ on the boundary of $V$. Define $\\phi_{3}=\\phi_{1}-\\phi_{2}$. By the linearity of Laplace's equation, $\\nabla^{2} \\phi_{2}=0$ and $\\phi_{3}=0$ on the surface. By the maximum principle, all extrema occurs at the surface. Thus, both the minimum and the maximum of $\\phi_{3}$ is 0 . Therefore, $\\phi_{3}=0$. This concludes the proof that $\\phi_{1}=\\phi_{2}$ identically.""]"		"5. Polarization and Oscillation 129 130In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.131 132Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.133Context question:134(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.135Context answer:136开放性回答137"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English1383	(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)	"[""The homogeneity of space implies the following:\n\n$$\nX\\left(x_{2}+h, t, v\\right)-X\\left(x_{2}, t, v\\right)=X\\left(x_{1}+h, t, v\\right)-X\\left(x_{1}, t, v\\right)\n$$\n\n\n\n<img_4548>\n\nFigure 3: The shaded area is the ruler's trajectory in spacetime. Any measurement of its length must intersect the shaded area.\n\n\n\nnow dividing by $h$ and sending $h \\rightarrow 0$ is the partial derivative, therefore\n\n$$\n\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{2}}=\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{1}}\n$$\n\nThe same method is repeated for the other variables.""]"		"4. Lorentz Boost139 140In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:141 142$$143\begin{aligned}144x^{\prime} & =x-v t \\145t^{\prime} & =t146\end{aligned}147$$148 149In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:150 151$$152\begin{aligned}153x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\154t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)155\end{aligned}156$$157 158In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.159 160The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,161 162$$163\begin{aligned}164x^{\prime} & =X(x, t, v) \\165t^{\prime} & =T(x, t, v)166\end{aligned}167$$168 169which we will refer to as the generalized boost."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English1704	(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)	['The isotropy of space implies\n\n$$\n\\begin{aligned}\nX(-x, t,-v) & =-x^{\\prime}=-X(x, t, v) \\\\\nT(-x, t,-v) & =T(x, t, v)\n\\end{aligned}\n$$\n\nand then plugging into (9) we see that\n\n$$\n\\begin{aligned}\n& A(-v)=A(v) \\\\\n& B(-v)=-B(v) \\\\\n& C(-v)=-C(v) \\\\\n& D(-v)=D(v)\n\\end{aligned}\n$$']		"4. Lorentz Boost171 172In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:173 174$$175\begin{aligned}176x^{\prime} & =x-v t \\177t^{\prime} & =t178\end{aligned}179$$180 181In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:182 183$$184\begin{aligned}185x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\186t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)187\end{aligned}188$$189 190In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.191 192The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,193 194$$195\begin{aligned}196x^{\prime} & =X(x, t, v) \\197t^{\prime} & =T(x, t, v)198\end{aligned}199$$200 201which we will refer to as the generalized boost.202Context question:203(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)204Context answer:205\boxed{证明题}206 207 208Extra Supplementary Reading Materials:209 210Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form211 212$$213\begin{aligned}214X(x, t, v) & =A(v) x+B(v) t \\215T(x, t, v) & =C(v) x+D(v) t216\end{aligned}217$$218 219where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English2205	"(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:221 222$$223\begin{aligned}224A(v)^{2}-B(v) C(v) & =1 \\225D(v)^{2}-B(v) C(v) & =1 \\226C(v)(A(v)-D(v)) & =0 \\227B(v)(A(v)-D(v)) & =0 .228\end{aligned}229$$230 231(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$."	['The principle of relativity implies that the coordinate transformation can be inverted such that\n\n$$\n\\begin{aligned}\nX\\left(x^{\\prime}, t^{\\prime},-v\\right) & =x \\\\\nT\\left(x^{\\prime}, t^{\\prime},-v\\right) & =t\n\\end{aligned}\n$$\n\ntherefore, because $X, T$ are linear, we can write the generalized boost in matrix form, and then using the even/oddness of $A, B, C, D$ derived previously\n\n$$\n\\left(\\begin{array}{ll}\nA(v) & B(v) \\\\\nC(v) & D(v)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -B(v) \\\\\n-C(v) & D(v)\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n1 & 0 \\\\\n0 & 1\n\\end{array}\\right)\n$$\n\nwhich is precisely the system of equations listed.']		"4. Lorentz Boost232 233In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:234 235$$236\begin{aligned}237x^{\prime} & =x-v t \\238t^{\prime} & =t239\end{aligned}240$$241 242In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:243 244$$245\begin{aligned}246x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\247t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)248\end{aligned}249$$250 251In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.252 253The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,254 255$$256\begin{aligned}257x^{\prime} & =X(x, t, v) \\258t^{\prime} & =T(x, t, v)259\end{aligned}260$$261 262which we will refer to as the generalized boost.263Context question:264(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)265Context answer:266\boxed{证明题}267 268 269Extra Supplementary Reading Materials:270 271Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form272 273$$274\begin{aligned}275X(x, t, v) & =A(v) x+B(v) t \\276T(x, t, v) & =C(v) x+D(v) t277\end{aligned}278$$279 280where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.281Context question:282(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)283Context answer:284\boxed{证明题}285"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English2866	"(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:287 288$$289\begin{aligned}290x^{\prime} & =A(v) x-v A(v) t \\291t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t292\end{aligned}293$$"	"[""The $F^{\\prime}$ frame is defined by $x^{\\prime}=0$, therefore $A(v) x+B(v) t=A(v) v t+B(v) t$ plugging in the equation $x=v t$. This implies that $A v+B=0$. This let's all the undetermined functions $A, B, C, D$ to be solved in terms of $A$.""]"		"4. Lorentz Boost294 295In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:296 297$$298\begin{aligned}299x^{\prime} & =x-v t \\300t^{\prime} & =t301\end{aligned}302$$303 304In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:305 306$$307\begin{aligned}308x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\309t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)310\end{aligned}311$$312 313In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.314 315The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,316 317$$318\begin{aligned}319x^{\prime} & =X(x, t, v) \\320t^{\prime} & =T(x, t, v)321\end{aligned}322$$323 324which we will refer to as the generalized boost.325Context question:326(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)327Context answer:328\boxed{证明题}329 330 331Extra Supplementary Reading Materials:332 333Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form334 335$$336\begin{aligned}337X(x, t, v) & =A(v) x+B(v) t \\338T(x, t, v) & =C(v) x+D(v) t339\end{aligned}340$$341 342where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.343Context question:344(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)345Context answer:346\boxed{证明题}347 348 349Context question:350(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:351 352$$353\begin{aligned}354A(v)^{2}-B(v) C(v) & =1 \\355D(v)^{2}-B(v) C(v) & =1 \\356C(v)(A(v)-D(v)) & =0 \\357B(v)(A(v)-D(v)) & =0 .358\end{aligned}359$$360 361(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.362Context answer:363\boxed{证明题}364"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English3657	"(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that366 367$$368\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .369$$370 371for an arbitrary constant $\kappa$."	['Consider a frame $F^{\\prime \\prime}$ related to $F^{\\prime}$ by a boost in the $x^{\\prime}$-direction with relative velocity $u$. Therefore, the composition of these two boosts results in a boost $F \\rightarrow F^{\\prime \\prime}$ given by\n\n$$\n\\Lambda_{F \\rightarrow F^{\\prime \\prime}}=\\left(\\begin{array}{cc}\nA(u) & -u A(u) \\\\\n-\\frac{A(u)^{2}-1}{u A(u)} & A(u)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -v A(v) \\\\\n-\\frac{A(v)^{2}-1}{v A(v)} & A(v)\n\\end{array}\\right)\n$$\n\nMultiplying out these matrices and noting that we have shown that the diagonal terms must be equal, we find that\n\n$$\n\\frac{A(v)^{2}-1}{v^{2} A(v)^{2}}=\\frac{A(u)^{2}-1}{u^{2} A(u)^{2}}\n$$\n\nas the left hand side is a function of $v$ only and the right hand side is a function of $u$ only, they must both be equal to some constant $\\kappa$.']		"4. Lorentz Boost372 373In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:374 375$$376\begin{aligned}377x^{\prime} & =x-v t \\378t^{\prime} & =t379\end{aligned}380$$381 382In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:383 384$$385\begin{aligned}386x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\387t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)388\end{aligned}389$$390 391In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.392 393The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,394 395$$396\begin{aligned}397x^{\prime} & =X(x, t, v) \\398t^{\prime} & =T(x, t, v)399\end{aligned}400$$401 402which we will refer to as the generalized boost.403Context question:404(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)405Context answer:406\boxed{证明题}407 408 409Extra Supplementary Reading Materials:410 411Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form412 413$$414\begin{aligned}415X(x, t, v) & =A(v) x+B(v) t \\416T(x, t, v) & =C(v) x+D(v) t417\end{aligned}418$$419 420where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.421Context question:422(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)423Context answer:424\boxed{证明题}425 426 427Context question:428(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:429 430$$431\begin{aligned}432A(v)^{2}-B(v) C(v) & =1 \\433D(v)^{2}-B(v) C(v) & =1 \\434C(v)(A(v)-D(v)) & =0 \\435B(v)(A(v)-D(v)) & =0 .436\end{aligned}437$$438 439(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.440Context answer:441\boxed{证明题}442 443 444Context question:445(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:446 447$$448\begin{aligned}449x^{\prime} & =A(v) x-v A(v) t \\450t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t451\end{aligned}452$$453Context answer:454\boxed{证明题}455"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English4568	"(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form457 458$$459\begin{aligned}460x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\461t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)462\end{aligned}463$$"	['Solve for $A$,\n\n$$\nA(v)=\\frac{1}{\\sqrt{1-\\kappa v^{2}}}\n$$\n\nand the dimensions of $\\kappa$ are determined by the restriction that $\\kappa v^{2}$ is being added to a dimensionless number. Substituting this form of $A(v)$ into\n\n$$\n\\begin{aligned}\nx^{\\prime} & =A(v) x-v A(v) t \\\\\nt^{\\prime} & =-\\left(\\frac{A(v)^{2}-1}{v A(v)}\\right) x+A(v) t\n\\end{aligned}\n$$\n\nyields the answer.']		"4. Lorentz Boost464 465In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:466 467$$468\begin{aligned}469x^{\prime} & =x-v t \\470t^{\prime} & =t471\end{aligned}472$$473 474In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:475 476$$477\begin{aligned}478x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\479t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)480\end{aligned}481$$482 483In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.484 485The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,486 487$$488\begin{aligned}489x^{\prime} & =X(x, t, v) \\490t^{\prime} & =T(x, t, v)491\end{aligned}492$$493 494which we will refer to as the generalized boost.495Context question:496(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)497Context answer:498\boxed{证明题}499 500 501Extra Supplementary Reading Materials:502 503Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form504 505$$506\begin{aligned}507X(x, t, v) & =A(v) x+B(v) t \\508T(x, t, v) & =C(v) x+D(v) t509\end{aligned}510$$511 512where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.513Context question:514(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)515Context answer:516\boxed{证明题}517 518 519Context question:520(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:521 522$$523\begin{aligned}524A(v)^{2}-B(v) C(v) & =1 \\525D(v)^{2}-B(v) C(v) & =1 \\526C(v)(A(v)-D(v)) & =0 \\527B(v)(A(v)-D(v)) & =0 .528\end{aligned}529$$530 531(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.532Context answer:533\boxed{证明题}534 535 536Context question:537(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:538 539$$540\begin{aligned}541x^{\prime} & =A(v) x-v A(v) t \\542t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t543\end{aligned}544$$545Context answer:546\boxed{证明题}547 548 549Context question:550(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that551 552$$553\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .554$$555 556for an arbitrary constant $\kappa$.557Context answer:558\boxed{证明题}559"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English5609	(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.	['If $v$ is unbounded and $\\kappa \\neq 0$, it may be large enough so that the square root gives an imaginary number, and as $x^{\\prime}, t^{\\prime}$ cannot be imaginary, it must be that $\\kappa=0$. There is nothing stopping a particle from traveling arbitrarily fast under a Galilean structure of spacetime, as given any particle we may Galilean boost to an inertial frame moving at $v$ arbitrarily fast, in which the particle is then moving at $-v$. By the principle of relativity, there is nothing wrong with doing physics in this frame, so it must be that it is acceptable to have particles move arbitrarily fast']		"4. Lorentz Boost561 562In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:563 564$$565\begin{aligned}566x^{\prime} & =x-v t \\567t^{\prime} & =t568\end{aligned}569$$570 571In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:572 573$$574\begin{aligned}575x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\576t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)577\end{aligned}578$$579 580In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.581 582The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,583 584$$585\begin{aligned}586x^{\prime} & =X(x, t, v) \\587t^{\prime} & =T(x, t, v)588\end{aligned}589$$590 591which we will refer to as the generalized boost.592Context question:593(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)594Context answer:595\boxed{证明题}596 597 598Extra Supplementary Reading Materials:599 600Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form601 602$$603\begin{aligned}604X(x, t, v) & =A(v) x+B(v) t \\605T(x, t, v) & =C(v) x+D(v) t606\end{aligned}607$$608 609where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.610Context question:611(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)612Context answer:613\boxed{证明题}614 615 616Context question:617(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:618 619$$620\begin{aligned}621A(v)^{2}-B(v) C(v) & =1 \\622D(v)^{2}-B(v) C(v) & =1 \\623C(v)(A(v)-D(v)) & =0 \\624B(v)(A(v)-D(v)) & =0 .625\end{aligned}626$$627 628(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.629Context answer:630\boxed{证明题}631 632 633Context question:634(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:635 636$$637\begin{aligned}638x^{\prime} & =A(v) x-v A(v) t \\639t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t640\end{aligned}641$$642Context answer:643\boxed{证明题}644 645 646Context question:647(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that648 649$$650\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .651$$652 653for an arbitrary constant $\kappa$.654Context answer:655\boxed{证明题}656 657 658Context question:659(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form660 661$$662\begin{aligned}663x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\664t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)665\end{aligned}666$$667Context answer:668\boxed{证明题}669"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English67010	(h) Assume that $v$ must be smaller than a finite value. Show that $1 / \sqrt{\kappa}$ is the maximum allowable speed, and that this speed is frame invariant, i.e., $\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x}{d t}$ for something moving at speed $1 / \sqrt{\kappa}$. Experiment has shown that this speed is $c$, the speed of light. Setting $\kappa=1 / c^{2}$, show that you recover the Lorentz boost.	['Again by requiring that $x^{\\prime}, t^{\\prime}$ are real, we can find the desired bound on $v$ from $1-\\kappa v^{2}>0$. One way to show that the speed is frame invariant is by deriving the relativistic velocity addition formula as follows\n\n$$\n\\begin{aligned}\nd x^{\\prime} & =\\gamma(d x-v d t) \\\\\nd t^{\\prime} & =\\gamma\\left(d t-v / c^{2} d x\\right)\n\\end{aligned}\n$$\n\nand dividing to yield\n\n$$\n\\frac{d x^{\\prime}}{d t^{\\prime}}=\\frac{\\frac{d x}{d t}-v}{1-\\frac{v}{c^{2}} \\frac{d x}{d t}}\n$$\n\nLet $w=d x^{\\prime} / d t^{\\prime}, u=d x / d t$, and, as often makes relativity problems easier to deal with, set $c=1$ (this is equivalent to choosing a new system of units). Then, we have\n\n$$\nw=\\frac{u-v}{1-v u}\n$$\n\nnow if $w=c=1$, we can solve the above equation to show that $u=1$, in other words frame $F$ and $F^{\\prime}$ both agree on what velocities move at $c$.']		"4. Lorentz Boost671 672In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:673 674$$675\begin{aligned}676x^{\prime} & =x-v t \\677t^{\prime} & =t678\end{aligned}679$$680 681In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:682 683$$684\begin{aligned}685x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\686t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)687\end{aligned}688$$689 690In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.691 692The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,693 694$$695\begin{aligned}696x^{\prime} & =X(x, t, v) \\697t^{\prime} & =T(x, t, v)698\end{aligned}699$$700 701which we will refer to as the generalized boost.702Context question:703(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)704Context answer:705\boxed{证明题}706 707 708Extra Supplementary Reading Materials:709 710Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form711 712$$713\begin{aligned}714X(x, t, v) & =A(v) x+B(v) t \\715T(x, t, v) & =C(v) x+D(v) t716\end{aligned}717$$718 719where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.720Context question:721(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)722Context answer:723\boxed{证明题}724 725 726Context question:727(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:728 729$$730\begin{aligned}731A(v)^{2}-B(v) C(v) & =1 \\732D(v)^{2}-B(v) C(v) & =1 \\733C(v)(A(v)-D(v)) & =0 \\734B(v)(A(v)-D(v)) & =0 .735\end{aligned}736$$737 738(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.739Context answer:740\boxed{证明题}741 742 743Context question:744(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:745 746$$747\begin{aligned}748x^{\prime} & =A(v) x-v A(v) t \\749t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t750\end{aligned}751$$752Context answer:753\boxed{证明题}754 755 756Context question:757(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that758 759$$760\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .761$$762 763for an arbitrary constant $\kappa$.764Context answer:765\boxed{证明题}766 767 768Context question:769(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form770 771$$772\begin{aligned}773x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\774t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)775\end{aligned}776$$777Context answer:778\boxed{证明题}779 780 781Context question:782(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.783Context answer:784\boxed{证明题}785"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English78611	"(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.787 788Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$."	['A periodic wave on interval $[0, L]$ must fit an integer number of wavelengths $\\lambda$ into the length $L$,\n\n$$\nn \\lambda=L\n\\tag{1}\n$$\n\nTherefore $k_{n}=\\frac{2 \\pi}{\\lambda_{n}}=\\frac{2 \\pi n}{L}$.\n\nFor the next part, use $c^{\\prime}=\\frac{\\omega}{k}$. Therefore,\n\n$$\n\\omega_{n}=c^{\\prime} k_{n}=\\frac{2 \\pi c^{\\prime} n}{L} \\Rightarrow d \\omega_{n}=\\frac{2 \\pi c^{\\prime}}{L} d n \\Rightarrow \\frac{d n}{d \\omega_{n}}=\\frac{L}{2 \\pi c^{\\prime}}\n\\tag{2}\n$$']		"2. Johnson-Nyquist noise789 790In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :791 792$$793\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R794\tag{2}795$$796 797Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.798 799Electromagnetic modes in a resistor800 801We first establish the properties of thermally excited electromagnetic modes802 803$$804V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)805\tag{3}806$$807 808in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$."	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English80912	"(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is810 811$$812\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}813\tag{4}814$$815 816In the low-energy limit $\hbar \omega_{n} \ll k T$, show that817 818$$819\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}820\tag{5}821$$822 823You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$."	['Use the given Taylor expansion,\n\n$$\n\\left\\langle N_{n}\\right\\rangle=\\frac{1}{\\exp \\frac{\\hbar \\omega_{n}}{k T}-1} \\approx \\frac{1}{\\left(1+\\frac{\\hbar \\omega_{n}}{k T}\\right)-1}=\\frac{k T}{\\hbar \\omega_{n}}\n\\tag{3}\n$$']		"2. Johnson-Nyquist noise824 825In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :826 827$$828\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R829\tag{2}830$$831 832Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.833 834Electromagnetic modes in a resistor835 836We first establish the properties of thermally excited electromagnetic modes837 838$$839V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)840\tag{3}841$$842 843in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.844Context question:845(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.846 847Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.848Context answer:849\boxed{证明题}850"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English85113	(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.	['The electromagnetic modes are excitations of the electromagnetic field like photons, so they are also massless. The energy of a photon is $E=h f=\\hbar \\omega$.\n\nAlternatively, you can start from $E^{2}=\\left(m c^{2}\\right)^{2}+(p c)^{2}$ and use $m=0, p=h / \\lambda$.']		"2. Johnson-Nyquist noise852 853In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :854 855$$856\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R857\tag{2}858$$859 860Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.861 862Electromagnetic modes in a resistor863 864We first establish the properties of thermally excited electromagnetic modes865 866$$867V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)868\tag{3}869$$870 871in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.872Context question:873(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.874 875Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.876Context answer:877\boxed{证明题}878 879 880Context question:881(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is882 883$$884\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}885\tag{4}886$$887 888In the low-energy limit $\hbar \omega_{n} \ll k T$, show that889 890$$891\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}892\tag{5}893$$894 895You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.896Context answer:897\boxed{证明题}898"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English89914	"(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is900 901$$902P[f, f+d f] \approx k T d f .903\tag{6}904$$905 906Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)"	['Power equals energy per time. The average energy delivered by one boson to the resistor is $\\hbar \\omega_{n}$.\n\nFor each state $n$, the number of bosons which either enter or exit the resistor per time is equal to their population divided by the time taken to travel the length of the resistor, $t=L / c^{\\prime}$ :\n\n$$\n\\frac{d\\left\\langle N_{n}\\right\\rangle}{d t}=\\frac{\\left\\langle N_{n}\\right\\rangle}{L / c^{\\prime}}=\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\n\\tag{4}\n$$\n\n\n\nFor a frequency interval $d \\omega_{n}$, the number of states is $d n=\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}$ (from part (a)). Therefore, the energy delivered per time, per frequency interval is\n\n$$\nd P\\left[\\omega_{n}, \\omega_{n}+d \\omega_{n}\\right]=\\left(\\hbar \\omega_{n}\\right) \\times\\left(\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\\right) \\times\\left(\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}\\right)=k T \\frac{d \\omega_{n}}{2 \\pi}\n\\tag{5}\n$$\n\nUsing $f=\\frac{\\omega}{2 \\pi}$,\n\n$$\nd P[f, f+d f]=k T d f\n\\tag{6}\n$$']		"2. Johnson-Nyquist noise907 908In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :909 910$$911\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R912\tag{2}913$$914 915Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.916 917Electromagnetic modes in a resistor918 919We first establish the properties of thermally excited electromagnetic modes920 921$$922V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)923\tag{3}924$$925 926in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.927Context question:928(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.929 930Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.931Context answer:932\boxed{证明题}933 934 935Context question:936(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is937 938$$939\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}940\tag{4}941$$942 943In the low-energy limit $\hbar \omega_{n} \ll k T$, show that944 945$$946\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}947\tag{5}948$$949 950You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.951Context answer:952\boxed{证明题}953 954 955Context question:956(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.957Context answer:958\boxed{证明题}959"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English96015	"(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is961 962$$963P_{\max }=\frac{V^{2}}{4 R} .964\tag{7}965$$966 967Give the optimal value of $r$ in terms of $R$ and $V$."	['The total resistance in this circuit is $R+r$. The power dissipated in $r$ is therefore\n\n$$\nP=r I^{2}=r\\left(\\frac{V}{R+r}\\right)^{2}\n\\tag{7}\n$$\n\nThe maximization of $\\frac{r}{(R+r)^{2}}$ is equivalent to the minimization of $\\phi(r)=\\frac{(R+r)^{2}}{r}=\\frac{R^{2}}{r}+2 R+r$. Setting $\\frac{d \\phi}{d r}=0$, for example, gives the solution $r=R$. Substituting $r=R$ into 7 yields\n\n$$\nP=\\frac{V^{2}}{4 R}\n\\tag{8}\n$$']		"2. Johnson-Nyquist noise968 969In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :970 971$$972\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R973\tag{2}974$$975 976Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.977 978Electromagnetic modes in a resistor979 980We first establish the properties of thermally excited electromagnetic modes981 982$$983V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)984\tag{3}985$$986 987in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.988Context question:989(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.990 991Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.992Context answer:993\boxed{证明题}994 995 996Context question:997(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is998 999$$1000\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}1001\tag{4}1002$$1003 1004In the low-energy limit $\hbar \omega_{n} \ll k T$, show that1005 1006$$1007\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}1008\tag{5}1009$$1010 1011You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.1012Context answer:1013\boxed{证明题}1014 1015 1016Context question:1017(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.1018Context answer:1019\boxed{证明题}1020 1021 1022Context question:1023(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is1024 1025$$1026P[f, f+d f] \approx k T d f .1027\tag{6}1028$$1029 1030Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)1031Context answer:1032\boxed{证明题}1033 1034 1035Extra Supplementary Reading Materials:1036 1037Nyquist equivalent noisy voltage source 1038 1039The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$."	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English104016	(b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$.	['Differentiate the result of the previous part, and apply a time-expectation $\\langle\\rangle: d\\left\\langle V^{2}\\right\\rangle=4 R d\\langle P\\rangle=$ $k T d f$. Therefore $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$.']		"2. Johnson-Nyquist noise1041 1042In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :1043 1044$$1045\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R1046\tag{2}1047$$1048 1049Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.1050 1051Electromagnetic modes in a resistor1052 1053We first establish the properties of thermally excited electromagnetic modes1054 1055$$1056V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)1057\tag{3}1058$$1059 1060in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.1061Context question:1062(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.1063 1064Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.1065Context answer:1066\boxed{证明题}1067 1068 1069Context question:1070(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is1071 1072$$1073\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}1074\tag{4}1075$$1076 1077In the low-energy limit $\hbar \omega_{n} \ll k T$, show that1078 1079$$1080\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}1081\tag{5}1082$$1083 1084You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.1085Context answer:1086\boxed{证明题}1087 1088 1089Context question:1090(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.1091Context answer:1092\boxed{证明题}1093 1094 1095Context question:1096(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is1097 1098$$1099P[f, f+d f] \approx k T d f .1100\tag{6}1101$$1102 1103Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)1104Context answer:1105\boxed{证明题}1106 1107 1108Extra Supplementary Reading Materials:1109 1110Nyquist equivalent noisy voltage source 1111 1112The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.1113Context question:1114(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is1115 1116$$1117P_{\max }=\frac{V^{2}}{4 R} .1118\tag{7}1119$$1120 1121Give the optimal value of $r$ in terms of $R$ and $V$.1122Context answer:1123证明题1124"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English112517	(a) Explain why no Johnson-Nyquist noise is produced by ideal inductors or capacitors. There are multiple explanations; any explanation will be accepted. (Hint: the impedance of an ideal inductor or capacitor is purely imaginary.)	"[""For example, the impedance of an ideal inductor or capacitor is purely imaginary. Replacing $R \\rightarrow i \\omega L, \\frac{1}{i \\omega C}$ in the formula $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$ would give an imaginary squared-voltage, which doesn't make sense.\n\nMore physically, the current and voltage in a pure inductor or capacitor are always orthogonal (out-of-phase), so no power is dissipated.""]"		"2. Johnson-Nyquist noise1126 1127In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :1128 1129$$1130\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R1131\tag{2}1132$$1133 1134Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.1135 1136Electromagnetic modes in a resistor1137 1138We first establish the properties of thermally excited electromagnetic modes1139 1140$$1141V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)1142\tag{3}1143$$1144 1145in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.1146Context question:1147(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.1148 1149Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.1150Context answer:1151\boxed{证明题}1152 1153 1154Context question:1155(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is1156 1157$$1158\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}1159\tag{4}1160$$1161 1162In the low-energy limit $\hbar \omega_{n} \ll k T$, show that1163 1164$$1165\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}1166\tag{5}1167$$1168 1169You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.1170Context answer:1171\boxed{证明题}1172 1173 1174Context question:1175(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.1176Context answer:1177\boxed{证明题}1178 1179 1180Context question:1181(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is1182 1183$$1184P[f, f+d f] \approx k T d f .1185\tag{6}1186$$1187 1188Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)1189Context answer:1190\boxed{证明题}1191 1192 1193Extra Supplementary Reading Materials:1194 1195Nyquist equivalent noisy voltage source 1196 1197The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.1198Context question:1199(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is1200 

Showing the first 1,200 of 1663 lines. Download the file for the rest.